N V Tejaswi

National Undergraduate Pro­

gramme in Mathematics and

Computer Science,

SPIC Mathematical Institute

92, G N Chetty Road

T Nagar,Chennai 600017 ,India.

N V Tejaswi studied in Ban­

galore (Vijaya High School and

the National College, Jaya­

nagar). He won a Gold Medal

at the International Mathe­

matical Olympiad held in

China-Taipei in July 1998.

After finishing his 2nd PUC in

1998 he is pursuing Mathe­

matics at the SPIC Mathe-

matical Institute, Chennai.

CLASSROOM

The Indian Gold Rush

I t was a great moment for Indians, when we got three gold and three silver medals in the 39th International Mathemat­ical Olympiad, held in Taipei, Taiwan in July '98. It was well organised in almost all aspects. First of all, the prob­lems in the final paper were of good standard. We had nice excursions, good cultural programmes, good guide,... etc., though the food situation was rather difficult for us, espe­cially the vegetarians amongst us (Rishi Raj, Yogananda and I survived on fruits).

Our perfonnance was really great. We once again proved our strength in GeOI:netry. India was the only country to get all six full 7 points in three problems. After the first day's examination was over in which all of us had got two problems right we were pleasantly surprised to hear our deputy leader, Yogananda, tell us that the second problem was a proposal from India! We, of course, did not know this. We are proud of R B Bapat, Indian Statistical Institute, Delhi, who was the author of this problem. More than 250 contestants got zero in that problem in which our team scored 42 out of 42! The one thing that is not in favour of us is that none of us could solve the 3rd and the 6th problem completely.

Here , I briefly discuss the problems which appeared in IMO:

Problem 1. In the convex quadrilateral ABCD, the diag­onals AC and B D are perpendicular and the opposite sides AB and DC are ' not parallel. Suppose that the point P, where the perpendicular bisectors of AB and DC meet, is inside ABC D. Prove that ABC D is a cyclic quadrilateml if and only if the triangles AB P and CD P have equal areas.

Our team found this problem rather easy and altogether gave five different solutions, two using co-ordinate geome­try, one using calculus and the other two being pure geom-

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CLASSROOM

etry solutions. If the quadrilateral is cyclic, then it is easy to see that the triangles AP Band C P D have same area. While proving the other way, I assumed that the result is not true and came to a contradiction. If the result is not true, then, assume AP > CP (since they are not equal). We draw a circle with P as centre and FA as radius. Then extend the diagonals to intersect the circle again at X and Y Thus triangle X PY = triangle AP B = triangle C P D (this follows frOln the previous implication), which leads to a contradiction.

Chetan and Soham solved by proving the fact that OM P N is a parallelogram, where 0 is the point of intersection of diagonals and M, N are midpoints of AB, CD respectively.

Problem 2. In a competition, there are a contestants and b judges, where b ~ 3 is an odd integer, Each judge rates each contestant a.s either pass or fail. Suppose k is a number such that, for any two judges, their Tatings coincide for at most k contestants. Prove that

k b-1 ->--a - 2b .

This is a problem which requires enumerating skill. And counting is the only way for solution. So all of us got similar solutions.

The main idea is counting the total number of agreements in two different ways, first via judges and then via contestants. Denote the judges by Ji, i = 1,2, ... , b and the contestants by Cr, r ::::: 1,2 ... , a. Let kij be the number of agreements judges J i and Jj have; there are (;) such kij's and since any two judges have atnlost k agreements the average of the kij's

is not greater than k, i.e., k ~ ill, On the other hand, if Pr

is the number of 'pass' got by Cr then the contribution to the total number of agreements by the Pr judges who 'pass' Or is (~); similarly for the number, a - Pr, of 'fails' got by Cr. Thus, the contribution to the total number of agreements

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CLASSROOM

by Cr is (~) + (U-;Pr-). Therefore we get,

a

E(P2") + (U2Pr )

k ~. ~i) == r=1 (~) The required inequality now follows from this by using the following result: let l be an integer and m + n == 2l + 1; then

Problem 3. For any positive integer n, let d(n) denote the number of positive divisors of n (including 1 and n itself).

Determine all positive integers k r5uch that

Jor some n.

d(n2

) == k d(n)

This is an interesting problem. Firstly, I noticed that k should be odd, and after getting first few values, I conjec­tured that k can take any odd value and it was true.

The proof was by induction. It was enough to show that each odd number k is expressible as

where ai ~ O.

(4al + 1)(4a2 + 1)· . (4al + 1) (2al + 1)(2az + 1)· . (2al + 1)'

It is easy to see that this function is multiplicative, i.e, if k can take values a and b, then k can take the value abo Hence it requires to prove only for primes. If the prime is of fonn P == 4m + 1, then we choose al = m and we prove that k can take the value p. If p is of form 4m + 3, then we write p = 2T S - 1 where s is odd and r ~ 2. Then we choose al =: (3p;l). Therefore, (2al + 1) == 3p;-1 =: PI (say). Then

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CLASSROOM

we take a2 = (3P~-1) and hence (2a2 + 1)- = (3P~+1) = P2.

Similarly we cho~e a3, a4, .. " ar. Then we get

(4al + 1)(4a2 + 1)· . (4ar + 1) P (2al + 1)(2a2 + 1) .. · (2ar + 1) == 28 + I'

But k can take the value 28 + 1 (by induction) and hence k can take the value p.

Problem 4. Determine all pairs (a, b) of positive integers such that ab2 + b + 7 divides a2b + a + b.

This was an easy problem. We have

b(a2b + a +b) = a(ab2 + b + 7) + (b2 - 7a),

Since ab2 + b + 7 divides L.H.S, we get ab2 + b + 7 divides b2 -7 a. If we consider three cases, (i)b2 - 7 a > 0,. (ii)b2 -7 a == 0, (iii)b2 - 7a < 0, then we get the solutions as (a, b) (11,1), (49,1), (7k 2 , 7k), where k ~ 1.

Problem 5. Let I be the incentre of triangle ABC. Let the incircle of ABC touch the sides BC, CA and AB at K, Land M, respectively. The line through B parallel to M K meets the lines LM and LK at Rand S, respectively. Prove that L R1 S is acute.

Since this is a geometry problem, we found this easy. Rishi Raj gave a wonderful proof using harmonic pencils. He took N as the midpoint of K M and proved that the triangle RN S is a self polar triangle w.r. t the incircle (i.e, R is the polar of N S, S is the polar of RN and N is the polar of RS). This implies that lis the orthocentre of triangle RNS in which LRN S is obtuse. Hence LRl S is acute.

I used the fact that S B M Land RB KLare cyclic. Hence I got RB· RS == RM RL, SE SR == SK SL. Therefore

(RJ2 -r2)+(SJ2 -r2)=RM RL+SK SL=

RB· RS+SR· BB = RS2•

Hence cosLRIS = RI~M)(sJ)2 > O. Thus LRIS is acute.

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CLASSROOM

Problem 6. Consider all junctions f /rom the set N of all positive integers into itself satisfying

1 (t2 1 (s )) = 8 (I {t))2 ,

for all sand t in N. Determine the least possible value of /(1998).

This is a good problem and is treated as the toughest prob­lem of this IMO. First, it is eaSy to see that /(/(8» = s/(1)2. Using this and manipulating the given equation we get that f(1)/(t2} = f(t)2. This implies that if prime p divides f(1} then p divides f(t) for all t ~ 1. It can be shown that 1(1) divides I(t) for all t ~ 1. Then we define g( t) == fg~, which .also satisfies the given conditions. We also have gg(x) = x and 9 is multiplicative. Therefore 9 takes primes to primes. Since 1998 = 2 33 37, we define g(3) = 2, g(37) = 5 which implies g(2) = 3, g(5) = 37 and g(p) == p for all primes p other than 2,3,5,37. Hence we get that the least value of g(1998) = 120. And hence the least value of f(1998) == 120.

I think it was a wonderful performance by the Indians and we are happy that we were able to offer our humble Golden Tribute to India in the Golden Jubilee year of her Indepen­dence. I thank each and everyone who is directly or indi­rectly involved in this effort. I hope that we get six golds next year.

Can a person be blamed

who is born with crabby

genes.

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